# Linear Inequalities

This chapter explores linear inequalities. Before attempting this chapter you must have prior knowledge of solving linear equations. The chapter covers solving linear inequalities, displaying the solutions on a number line, solving double-ended inequalities, and look at the intersection of two solutions.

## Equations and inequalities connection

Example:

Answer:

24 – x = 3(2x + 1)

24 – x = 6x + 3

24 + 3 = 6x + x

21 = 7x

x = 3

Explanation: You must already know how to solve linear equations for example below is an equation to solve naturally.

24 – x = 3(2x + 1)

First we expand the equation as shown below.24 – x = 6x + 3

Arrange x terms on one side as shown below.24 + 3 = 6x + x

Now we can simplify as shown below21 = 7x

We divide both sides by 7 and the result becomes.x = 3

Example:

Answer:

24 – x ≤ 3(2x + 1)

24 – x ≤ 6x + 3

24 – 3 ≤ 6x + x

21 ≤ 7x

3 ≤ x

x ≤ 3

Explanation: We can use the same principle used above to solve the following inequality. We can also indicate this on a number line to get a better visual as shown below.

24 – x ≤ 3(2x + 1)

First we expand as shown below.24 – x ≤ 6x + 3

Rearrange24 – 3 ≤ 6x + x

…and simplify…21 ≤ 7x

…then divide both sides by 7 to simplify again. We get that;3 ≤ x

x ≤ 3

## Negative coefficient of x

Example: You may have a negative number of xs. For example below is an inequality that we could solve.

16 – 4x ≤ 22

Answer:

16 < 22 + 4x

16 – 22 ≤ 4x

-6 ≤ 4x

-1.5 ≤ x

x ≥ -1.5

-4x < 22 – 16

-4x ≤ 6

x ≥ -1.

Explanation: There are two ways we could solve the equality; we could move the x term to the right to make it positive as shown below. We divide both sides by -4 and the result becomes;

16 < 22 + 4x

And then move the integer to the other side as shown below.16 – 22 ≤ 4x

The result becomes;-6 ≤ 4x

We divide both sides by 4 and the result becomes;-1.5 ≤ x

orx ≥ -1.5

The other alternative is not moving the term with x and move 16 instead.-4x < 22 – 16

-4x ≤ 6

x ≥ -1.

The solution can also be shown on a number line as shown below. You must have noticed that when you divide or multiply by a negative, the inequality changes direction.## Double-ended inequalities

Example:Below is what a double-ended inequality looks like.

6 – x ≤ 2x + 3 < 15

Answer:

6 – x ≤ 2x + 3 2x + 3 < 15

6 – x ≤ 2x + 3 2x < 15 – 3

3 ≤ 3x 2x < 12

1 ≤ x x < 6

1 ≤ x < 6

Explanation: There are two parts to the inequality so we split it into two parts as shown below. Now we can put the solution back together to result.

6 – x ≤ 2x + 3 2x + 3 < 15

Now we can solve each part separately as have been done below.6 – x ≤ 2x + 3 2x < 15 – 3

3 ≤ 3x 2x < 12

1 ≤ x x < 6

1 ≤ x < 6

The solution has been shown on the number line below. ## Solving inequalities in a pair

Example: You could also work out where two solutions overlap. For example suppose we had to find the values which satisfy both inequalities shown below.

3x + 2 > x – 6 and 2x > 5x – 9

Answer:

3x + 2 > x – 6

3x – x > -6 – 2

2x > -8

x > -4

2x > 5x – 9

2x – 5x > – 9

-3x > -9

x > 3

x > -4 and x < 3

## Exam question

Example:ABC is a triangle with perimeter greater than 28cm and less than 36cm. The length of AB is an integer. AC is 2cm longer than AB. BC is 5cm shorter than AB.

Answer:

Perimeter = x + (x + 2) + (x – 5)

= 3x – 3

28 < 3x – 3 < 36

28 < 3x – 3 3x – 3 < 36

31 < 3x 3x < 39

101/3 < x < 13

101/3 < x < 13

x = 11, 12

Explanation:Find all possible values for the length of AB. First we find an expression for the perimeter. The perimeter is greater than 28cm and less than 36cm. The range of values x can take are;

Perimeter = x + (x + 2) + (x – 5)

= 3x – 3

28 < 3x – 3 < 36

The inequality contains two parts so we solve each part separately.28 < 3x – 3 3x – 3 < 36

31 < 3x 3x < 39

101/3 < x < 13

101/3 < x < 13

We know that AB is an integer so we look for whole numbers in the range that is;x = 11, 12

The length of AB is either 11cm or 12cm. 13 has not been included in the final answer. LINK TO PAGE

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